\(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(\frac{2}{5}.7+\frac{2}{7}.9+\frac{2}{9}.11+...+\frac{2}{49}.51\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\frac{46}{255}\)
\(=\frac{23}{255}\)
\(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.51}\)
\(\Rightarrow2 \left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\right)\)
\(\Rightarrow\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{51}=\frac{46}{255}\)
Vì biểu thức đã được nhân 2 nên giá trị của biểu thức là:
\(\frac{46}{255}:2=\frac{23}{255}\)
\(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\cdot\frac{46}{255}\)
\(=\frac{23}{255}\)
Đặt A .
\(A=\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\)
\(2A=2\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\right)\)
\(2A=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{49}-\frac{1}{51}\right)\)\(2A=2\left(\frac{1}{5}-\frac{1}{51}\right)\)
\(2A=\frac{2.46}{255}\)
\(A=\frac{2.46}{255}\)
\(A=\frac{92}{255}\)
Đặt dãy phép tính trên là A.
A= \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\)
\(2A=2\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\right)\)
\(2A=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...\frac{1}{49}-\frac{1}{51}\right)\)
\(2A=2\left(\frac{1}{5}-\frac{1}{51}\right)\)
\(2A=\frac{2.46}{255}\)
\(A=\frac{2.46}{255}\)
\(A=\frac{92}{255}\)